| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise perimeter or area of 2D region |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with guided steps. Part (a) requires setting up area equation and rearranging (algebraic manipulation), parts (b-c) use routine differentiation to find and verify a minimum, and part (d) involves simplifying the answer. The constraint is given, the perimeter formula is provided, and all steps follow a predictable textbook pattern with no novel insight required. Slightly easier than average due to heavy scaffolding. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{area} = 2xy + (\frac{1}{2} \times x^2 \times 0.5) = 2xy + \frac{1}{4}x^2 = 50\) | M1 | |
| \(\therefore y = \frac{50 - \frac{1}{4}x^2}{2x} = \frac{25}{x} - \frac{1}{8}x\) | A1 | |
| \(P = 2x + 4y + (x \times 0.5) = \frac{5}{2}x + 4y\) | M1 | |
| \(= \frac{5}{2}x + 4(\frac{25}{x} - \frac{1}{8}x)\) | M1 | |
| \(= \frac{5}{2}x + \frac{100}{x} - \frac{1}{2}x = 2x + \frac{100}{x}\) | A1 | |
| (b) \(\frac{dP}{dx} = 2 - 100x^{-2}\) | M1 A1 | |
| \(\text{for minimum}, \quad 2 - 100x^{-2} = 0\) | M1 | |
| \(x^2 = 50\) | M1 | |
| \(x = \sqrt{50} \text{ or } 5\sqrt{2}\) | A1 | |
| (c) \(\frac{d^2P}{dx^2} = 200x^{-3}\) | M1 | |
| \(\text{when } x = 5\sqrt{2}, \quad \frac{d^2P}{dx^2} = \frac{2}{5}\sqrt{2}\), \(\frac{d^2P}{dx^2} > 0 \therefore \text{minimum}\) | A1 | |
| (d) \(= 2(5\sqrt{2}) + \frac{100}{5\sqrt{2}} = 10\sqrt{2} + 10\sqrt{2} = 20\sqrt{2}\) | M1 A1 | (13) |
(a) $\text{area} = 2xy + (\frac{1}{2} \times x^2 \times 0.5) = 2xy + \frac{1}{4}x^2 = 50$ | M1 |
$\therefore y = \frac{50 - \frac{1}{4}x^2}{2x} = \frac{25}{x} - \frac{1}{8}x$ | A1 |
$P = 2x + 4y + (x \times 0.5) = \frac{5}{2}x + 4y$ | M1 |
$= \frac{5}{2}x + 4(\frac{25}{x} - \frac{1}{8}x)$ | M1 |
$= \frac{5}{2}x + \frac{100}{x} - \frac{1}{2}x = 2x + \frac{100}{x}$ | A1 |
(b) $\frac{dP}{dx} = 2 - 100x^{-2}$ | M1 A1 |
$\text{for minimum}, \quad 2 - 100x^{-2} = 0$ | M1 |
$x^2 = 50$ | M1 |
$x = \sqrt{50} \text{ or } 5\sqrt{2}$ | A1 |
(c) $\frac{d^2P}{dx^2} = 200x^{-3}$ | M1 |
$\text{when } x = 5\sqrt{2}, \quad \frac{d^2P}{dx^2} = \frac{2}{5}\sqrt{2}$, $\frac{d^2P}{dx^2} > 0 \therefore \text{minimum}$ | A1 |
(d) $= 2(5\sqrt{2}) + \frac{100}{5\sqrt{2}} = 10\sqrt{2} + 10\sqrt{2} = 20\sqrt{2}$ | M1 A1 | (13)9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c7c8cf84-06ac-4059-b8f0-d68b6d1d8dcc-4_661_915_932_431}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a design consisting of two rectangles measuring $x \mathrm {~cm}$ by $y \mathrm {~cm}$ joined to a circular sector of radius $x \mathrm {~cm}$ and angle 0.5 radians.
Given that the area of the design is $50 \mathrm {~cm} ^ { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item show that the perimeter, $P$ cm, of the design is given by
$$P = 2 x + \frac { 100 } { x }$$
\item Find the value of $x$ for which $P$ is a minimum.
\item Show that $P$ is a minimum for this value of $x$.
\item Find the minimum value of $P$ in the form $k \sqrt { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [13]}}